1. Writing and Balancing Equations
What do we know already??

2. WORD EQUATION
Not much information other than the elements/compounds involved
Potassium metal + oxygen gas  potassium oxide
Products are on the right-hand side of the equation
Reactants are on the left-hand side of the equation
+ sign means “reacts with”
Arrow means produces/yields/ makes

3. SKELETON (Unbalanced) EQUATION
A skeleton equation shows the formulas of the elements/compounds – it is unbalanced
Shows atoms, but not quantities of atoms
K(s) + O2 (g)  K2O(s)

4. Law of Conservation of Mass!
Because of the Law of Conservation of Mass, we know that the reactants have to equal the products. So to balance an equation – there must be the same amount of atoms (for each element) on the reactant side and the product side
Word equation:
Methane + oxygen  water + carbon dioxide
Skeleton equation:
CH4(g) + O2 (g)  H2O( ) + CO2(g)
So we have to get the same number of C, H and O, on each side of the equation. (we can only add coefficients to do this!)

5. BALANCED EQUATION 4K(s) + O2 (g)  2K2O(s)
A balanced chemical equation shows all atoms and their quantities
Balancing ensures that the number of each atom is the same on both sides of the reaction arrow
Always use the smallest whole number ratio
You can only ever add Coefficients (in front)– not subscripts behind!
Hint, balance H’s 2nd last and O’s Last!!!
4K(s) + O2 (g)  2K2O(s)

6. Method #1: Counting Atoms by splitting the sides of the Equation2
Ba H2O  Ba(OH) H2
Ba ___
H ___
O ___ 1
Ba ___
H ___
O ___ 1 2
X 2=4 4 1
X 2=2 2
So to balance everything, we need 2 more I’s on the reactant side, so place a 2 in front of the H
We need 2 more O’s on the reactant side, so place a 2 in front of the O
WAIT!!!! There is already a 2 in front of the O from when we placed it to balance the H’s…great… it works out more easily than we thought.

7. Try another 3 2 3 2 1 1 1 3 1 Fe ___ Fe ___ C ___ C ___ O ___ O ___
Fe2O C  Fe CO
Fe ___
C ___
O ___ 2
Fe ___
C ___
O ___ 1
X 2=2 1 1 3 1
X 3=3
WAIT!!!! Now the C’s aren’t balanced anymore! We need 3 more C’s on the reactant side to balance it.

8. Try a harder one! 2 2 1 1 2 1 1 1 1 2 Pb ___ Pb ___ NO3 ___ NO3 ___
Pb(NO3) KI  PbI KNO3 2
Pb ___
NO3 ___
K ___
I ___ 1
Pb ___
NO3 ___
K ___
I ___ 1 2 1
X 2=2 1 1 1
X 2=2 2
WAIT!!!! That changed the K’s… but we were lucky… It balanced them both out to 2 each.

9. Try the hardest one! 2 7 4 6 3 C2H6 + O2  CO2 + H2O 2 2 1 6 2 2 3 7
3.5 2 2 1
X 2=2
C ___
H ___
O ___
C ___
H ___
O ___ 6 2
X 3=6 2
X 3.5=7 3 7
WAIT!!!! We can’t have a decimal in an equation.
So multiply everything by 2 to get the next whole number
2C2H O2  4CO H2O
Balanced Equation

10. Method #2: Counting Atoms to Balance an Equation
Because of the Law of Conservation of Mass, we can count atoms and use math to balance the number of atoms in chemical equations.
Word equation:
Methane + oxygen  water + carbon dioxide
Skeleton equation:
CH4(g) + O2 (g)  H2O( ) + CO2(g)
To balance the compounds, take note of how many atoms of each element are on each side of the reaction arrow. Try to make the same number of each element on each side!

11. CH4(g) + O2 (g)  H2O(aq ) + CO2(g)
1 Carbon, 4 Hydrogen, 2 Oxygen  Carbon, 2 Hydrogen, 3 Oxygen
Remember – do O last and H 2nd last.
To balance, attempt to find values that equate atoms on both sides

12. Balanced equation: CH4(g) + 2O2 (g)  2H2O(aq) + CO2(g)
1 Carbon, 4 Hydrogen, (2x2) Oxygen  Carbon, (2x2) Hydrogen, (2)+2 Oxygen
Remember, you can only add numbers as Coefficients ( #’s in front), never as subscripts behind!!!
Hint, balance H’s 2nd last and O’s Last!!!

13. TIPS for Balancing Equations
Balance chemical equations by following these steps:
Trial and error will work, but can be very inefficient
Use the table method
Polyatomic ions (such as SO42–) can often be balanced as a whole group
Only add coefficients (the #’s in front of the elements); NEVER change or add subscripts (behind the elements)!
Balance the H’s 2nd last
Balance O’s Last
Always double-check after you think you are finished!

14. Balance the following:
Fe + Br2  FeBr3
Sn(NO2)4 + K3PO4  KNO2 + Sn3(PO4)4
C2H4 + O2  CO2 + H2O
2Fe + 3Br2  2FeBr3
3Sn(NO2)4 + 4K3PO4  12KNO2 + Sn3 (PO4)4
C2H4 + 3O2  2CO2 + 2H2O